Liquid-Liquid Extraction (LLX)

Liquid-Liquid Extraction (LLX)

Extraction is a liquid-liquid operation. It is a process of transferring a solute from one

liquid phase to another immiscible or partially miscible liquid in contact with the first. The two phases are chemically quite different, which leads to a separation of the components according to their distribution or partition between the two phases, normally one organic and one water. This is different from distillation, in which the liquid is partially vaporized to create another (vapor) phase, but the two phases are similar chemically.

Equilibrium Relations in Extraction

1. Phase rule.

In a liquid-liquid system, the minimum number of components is three and we have two phases in equilibrium. For a ternary system, the number of degrees of freedom is three, as calculated by the phase rule (F = C - P + 2 = 3 - 2 + 2 = 3). We have six variables: temperature, pressure, and four concentrations. If the pressure and temperature are specified, then setting one concentration will fix the system. The other three concentrations must be calculated from the phase equilibrium.

2. Triangular coordinates and equilibrium data.

Since we have three components, the equilibrium data are often expressed in equilateral triangular coordinates. This is shown in the following figure.

The three corners represent three pure components, A, B and C. The point M is a mixture. The perpendicular distance from the point M to any base line is the mass fraction of the component at the corner opposite to that base line. For example, the distance from M to base AB is the mass fraction of C (xC = 0.40).

A com Fig 2 misci The tw envel separ equili also p which one p Figur The curve exper and S we c mixtu point mixtu mmon p , where ble, an wo pha lope. A ate into ibrium plotted h is cal phase re re 2. Li co equili e is riments S are can ch ure liqu D in ure is s phase d e a pai nd liqu ase regi An ori o two c tie line d. The lled the egion. iquid-li ompone ibrium obtaine s. If com partiall hoose uid of B n Fig. separat diagram ir of co id C d ion is i iginal m conjuga e throu e two p e Plait iquid p ents A a solu ed thr mponen ly mis any b B and S 3. ed into m is Typ ompone dissolve include mixture ate pha ugh poi phases point. phase di and B a ubility rough nts B cible, binary S, say This o two ype I sy ents A es com d insid e of co ases a a int M. are id Outsi iagram are par ystem a and B mpletely de below omposi and b w Other dentical ide the m where rtially m and sho are par y in A w the c ition M which a tie line l at po envelo e miscible wn in rtially or B. curved M will are on es are int P, ope is e.

Under constant temperature, if we add component A into this binary mixture drop by drop, the composition of the ternary mixture will change along line DA. The ratio of B/S is constant while the amount of A is changing. When the amount of A is just to make the mixture from two phases to one single homogeneous phase, the composition is represented by the point D’. By repeating this procedure for other binary mixtures with compositions of E, F, G, we obtain the points E’, F’, G’. The curve links the points PD’E’F’G’Q is the equilibrium solubility curve. If B and S are completely immiscible, the two end points P and Q on the solubility curve will superimpose to the top points B and S, respectively.

Conjugate line. As far as some equilibrium tie lines are

available, other tie lines can be obtained by interpolation with the help of conjugate curve. Assuming the tie lines E1R1, E2R2, E3R3, E4R4 are known, we draw a vertical line from E1, which

intersects the horizontal line from

R1 at point F. Similary, the vertical lines from E2, E3, E4,

intersects the horizontal lines from

R2, R3, R4 at points G, H and J. The curve joining FGHJ and the plait point P is called the conjugate line.

Example E1: Ternary acetic acid-benzene-water mixture.

The liquid-liquid equilibrium data at 25oC is given in the following table. In the right-angled triangle, illustrate

(1) Solubility curve

(2) Tie lines for experiments of Nos. 2, 3, 4, 6, 8 (3) Plait point and conjugate line

No. Benzene phase (mass %) Water phase (mass %) _acetic acid

benzene water acetic acid benzene water 1 0.15 99.85 0.001 4.56 0.04 95.4 2 1.4 98.56 0.04 17.7 0.20 82.1 3 3.27 96.62 0.11 29.0 0.40 70.6 4 13.3 86.3 0.4 56.9 3.3 39.8 5 15.0 84.5 0.5 59.2 4.0 36.8 6 19.9 79.4 0.7 63.9 6.5 29.6 7 22.8 76.35 0.85 64.8 7.7 27.5 8 31.0 67.1 1.9 65.8 18.1 16.1 9 35.3 62.2 2.5 64.5 21.1 14.4 10 37.8 59.2 3.0 63.4 23.4 13.2 11 44.7 50.7 4.6 59.3 30.0 10.7 12 52.3 40.5 7.2 52.3 40.5 7.2 (1)The data points are plotted in the right-angled triangle coordinates, joining the points gives the solubility curve. (2) The experiments of Nos. 2, 3, 4, 6, 8 are shown as points R1, E1, R2, E2, R3, E3, R4, E4, R5, E5. The tie lines are the straight lines R1E1, R2E2, R3E3, R4E4, R5E5.

(3) The last set of data have the same composition in the two phases, which is the plait point. The auxiliary curve is obtained by drawing vertical lines from E1, E2, E3, E4, E5, which intersect the horizontal lines from R1, R2, R3, R4, R5 at points G, H, I, J, L. Joining GHIJLP gives the conjugate line.

Example E2: For the ternary system of example E1 at

25oC, a mixture is separated into two liquid phases after settling. One phase contains 15% acetic acid, 0.5% water and the rest being benzene (all mass %). Use the conjugate line in example E1 to determine the composition of the other conjugate liquid phase and draw the tie line.

Solution: The solubility and conjugate curves are given.

1. Find the composition of 15% acetic acid, 0.5% water as point R.

2. Draw a horizontal line from R to intersect the conjugate line at Q.

3. Draw a vertical line from Q to intersect the solubility curve at E, which is the composition of the other conjugate phase, 59% acetic acid, 37% water, 4% benzene.

3. Equilibrium data on rectangular coordinates.

Because of the special coordinates, the triangular diagram is not convenient. The liquid-liquid phase equilibrium is more often presented in rectangular coordinates, which is shown in Fig. 4 for acetic acid (A) - water (B) - isopropyl ether solvent (C).

Figure 4. Acetic acid (A)-Water (B)-Isopropyl ether (C) liquid-liquid phase diagram at 293 K (20oC).

The solvent pair B and C are partially miscible. The concentration of A is plotted on the horizontal axis and that of C on the vertical axis. The concentration of B is calculated from the following equation

x_B =1 0. −x_A −x_C y_B =1 0. −y_A −y_C

A tie line gi is shown connecting the water-rich layer i, called the raffinate layer, and the ether-rich solvent layer g, called the extract layer. The raffinate composition is designated by x, and the extract by y. Hence, the mass fraction of C is designated as yC in the extract laywer and as xC in the raffinate layer. To construct the tie line gi using the equilibrium yA-xA plot below the phase diagram, vertical lines to g and i are drawn.

Example E3: Material balance for equilibrium layers

An original mixture weighing 100 kg and containing 30 kg of isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is equilibrated and the equilibrium phases separated. What are the compositions of the two equilibrium phases.

Solution: The composition of the original mixture is

xC = 0.30, xA = 0.10, xB = 0.60

This composition is plotted as point h on Fig. 4. The tie line gi is obtained through point h by trial and error. The composition of the extract (ether) layer at g is yA = 0.04, yC = 0.94, and yB = 1 – 0.04 – 0.94 = 0.02 mass fraction. The raffinate (water) layer composition at i is xA = 0.12, xC = 0.02, and xB = 1 – 0.12 – 0.02 = 0.86 mass fraction.

In the above system (Fig. 4) the solvent pair B and C are partially miscible, while A is totally soluble in B or C. It is also common for some other systems that both pairs, A and C, and B and C are partially miscible. This is Type II system and shown in figure 5.

Figure 5. Liquid-liquid phase diagram where the solvent pairs A-C and B-C are partially miscible. Examples are the systems of Styrene (A) Ethylbene (B) Diethylene Glycol (C), and Chlorobenzene (A) -Methylethyl Ketone (B) -Water (C).

Single-Stage Equilibrium Extraction

In an extraction process we have two entering streams (L kg and V kg) which are NOT in equilibrium, as shown in Fig. 6. The solvent, as stream V₂, enters and the stream L₀ enters from the other side. The two entering streams are mixed and equilibrated and then exit as streams L₁ and V₁, which are in equilibrium with each other. To find the final product compositions in the two phases, it is required to know the mixture total mass and composition (point M). This can be obtained by material balances. After the point M is identified, the product composition can be found by the equilibrium tie line.

Figure 6. Single-stage Liquid-liquid extraction: (a) process flow diagram, (b) phase diagram. Material balances:

Overall: L₀ +V₂ = L₁+V₁ = M (1) (A): L x_{0 A0} +V y_{2 A2} =L x_{1 A1}+V y_{1 A1} = Mx_AM (2) (C): L x_{0 C0} +V y_{2 C2} =L x_{1 C1}+V y_{1 C1} = Mx_CM (3) Since x_A + x_B + x_C = 1 , an equation for B is not needed. because L₀ and V₂ are known, values of M, x_AM, and x_CM, can be found from Eqs. (1) to (3). L₁ and V₁ are obtained by drawing a tie line through point M.

Derivation of lever-arm rule for graphical addition.

In Figure 7 we have two streams (L & V) mixed to give a resulting mixture stream M kg total mass.

((b)

(b)

Figure 7. Graphical addition and lever-arm rule: (a) process flow, (b) graphical addition. By doing material balances, we have

Overall: V L+ = M (4)

(A): Lx_A +Vy_A == Mx_AM (5)

(C): Lx_C+Vy_C == Mx_CM (6)

Combining Eqs. (4) & (5), and (4) & (6), we have L V x y x x AM A A AM = − − (7) L V x y x x CM C C CM = − − (8)

Equating Eqs. (7) & (8) and rearranging,

x x x x x y x y C CM A AM CM C AM A − − = − − (9)

The left side is the slope of line LM and the right side is the slope of line MV. Because the two slopes are the same and the two lines have a common point M, the three

points L, M, and V must be on a straight line. The lever-arm rule is L V VM LM = & L M VM LV = (10)

Example E4: Amount of phases in solvent extraction

The compositions of the two equilibrium layers in example E1 are:

for the extract layer (V),

yA = 0.04, yB = 0.02, yC = 0.94 for the raffinate layer (L),

xA = 0.12, xB = 0.86, xC = 0.02

The original mixture contained 100 kg and xAM = 0.10. Determine the amounts of V and L.

Solution: The overall material balance is

V + L = M = 100 kg The material balance of A is

V(0.04) + L(0.12) = 100(0.10) Hence,

V = 75 kg, L = 25 kg

Alternatively, using the lever-arm rule, the distance hg in Fig. 4 is measured as 4.2 units and gi as 5.8 units. Then

L M L 100 hg gı 4.2 5.8

Solving, L = 72.5 kg and V = 27.5 kg, which is in reasonable close agreement with the material balance method.

1. Countercurrent process and overall balances

A countercurrent multistage process is shown in Fig. 8.

Figure 8. Countercurrent multistage extraction process flow diagram

The overall balance on all N stages is

L₀ +V_N+1 =L_N+V₁ =M (11) where M is the total mass (kg/h) and is a constant, L0 the inlet feed flow rate (kg/h), VN+1 the inlet solvent flow rate (kg/h), V1 the exit extract stream, and LN the exit raffinate stream. Material balance on C gives

L x_{0 C0} +V_N+1y_{C N, +1} = L x_{N C N,} +V y_{1 C1} = Mx_CM (12) x_CM is obtained By solving Eqs. (11) & (12)

x L x V y L V L x V y L V CM C N C N N N CN C N = + + = + + + + + 0 0 1 1 0 1 1 1 1 , ₍₁₃₎

A similar balance on component A gives

x L x V y L V L x V y L V AM A N A N N N AN A N = + + = + + + + + 0 0 1 1 0 1 1 1 1 , ₍₁₄₎

So the point M, which ties together the two entering streams (usually known) and the two exit streams, can be located. The desired exit composition xAN is often set, which is on the equilibrium curve (phase boundary). Then the line L_NM is extended to intersect the phase boundary of the extract phase to give V1 composition.

Example E5: Pure solvent isopropyl ether (C) at the rate

of V_N+1 = 600 kg/h is being used to extract an aqueous solution of L₀ = 200 kg/h containing 30 wt % acetic acid (A) and 70 wt % water (B) by countercurrent multistage extraction. The desired exit acetic acid concentration in the aqueous phase is 4%. Calculate the compositions and amounts of the ether extract V₁ and the aqueous raffinate L_N. The equilibrium data at 20oC, 1 atm, are given and plotted below.

Water phase (mass fraction)

isopropyl ether phase (mass fraction) acetic acid (xA) Water (xB) isopropyl ether (xC) acetic acid (yA) Water (yB) isopropyl ether (yC) 6.9e-3 0.9810 0.0120 1.8e-3 5.0e-3 0.9930 0.0141 0.9710 0.0150 3.7e-3 7.0e-3 0.9890 0.0289 0.9550 0.0160 7.9e-3 8.0e-3 0.9840 0.0642 0.9170 0.0190 0.0193 0.0100 0.9710 0.1330 0.8440 0.0230 0.0482 0.0190 0.9330 0.2550 0.7110 0.0340 0.1140 0.0390 0.8470 0.3670 0.5890 0.0440 0.2160 0.0690 0.7150 0.4430 0.4510 0.1060 0.3110 0.1080 0.5810 0.4640 0.3710 0.1650 0.3620 0.1510 0.4870

Solution:

The given values are Pure solvent inlet:

V_N+1 = 600, y_A,N+1 = y_B,N+1 = 0, y_C,N+1 = 1, Feed:

L₀ = 200, x_A0 = 0.3, x_B0 = 0.7, x_C0 = 0, Raffinate:

xAN = 0.04.

VN+1 and L0 are located by the compositions.

Since LN is on the phase boundary of the raffinate phase,

The composition of the mixture, x_CM and x_AM, are calculated by Eqs. (13) & (14) as 0.75 & 0.075 and used to plot point M. V₁ is located by drawing a line from L_N through M & extending it until it intersects the phase boundary in the extract phase. This gives y_A1 = 0.08 & yC1 = 0.90.

By solving Eqs. (11) & (12), L_N = 136 kg/h & V1 = 664 kg/h.

2.Stage-to-stage calculation for countercurrent extraction The next step is to go stage by stage to determine the concentrations at each stage and the total number of stages N needed to reach LN in the process.

Making a total balance on stage 1 and then on stage n,

L

+

V

=

L

+

V

L

+

V

=

L

+

V

L

−

V

=

L

−

V

=

...

=

L

−

V

=

L

−

V

=

Δ

(17) The value of Δ is constant for all stages. The coordinates

L x_{0 A0} −V y_{1 A1} = ... =L x_{N AN} −V_N+1y_{A N, +1} =Δx_AΔ (18) L x_{0 C0} −V y_{1 C1} = ... =L x_{N CN} −V_N+1y_{C N, +1} =Δx_CΔ (19) x L x V y L V L x V y L V A A A N AN N A N N N Δ = ₋− = −₋ + + + 0 0 1 1 0 1 1 1 1 , ₍₂₀₎

Similar Eqs. for x_BΔ & x_CΔ can be obtained. This point Δ is located either by its coordinates as calculated by Eq. (20) or graphically as the intersection of lines L0V1 and LNVN+1. The method to locate V1 has been discussed in example E5. All the operating lines (L0V1, L1V2, LnVn+1, ... , LNVN+1) must pass through the common point

Δ.

To graphically determine the number of stages, follow the procedures below.

(1) locate L₀, V_N+1 and L_N by their compositions.

(2) draw a line L₀V_N+1, and locate the mixture point M by Eq. (13) or (14).

(3) draw a line from L_N through M & extend it until it intersects the phase boundary, where is V1.

(3) extend lines L0V1, and LNVN+1, which will intersect at the common operating point Δ.

(4) start at L₀ and draw a line L₀Δ which intersects the phase boundary at V₁.

(5) draw an equilibrium tie line through V₁ to locate L₁. (6) draw a line L1Δ to give V2 at the phase boundary. (7) a tie line from V₂ gives L₂. This is continued until the desired LN is reached.

Alternately, the point Δ can firstly be located using Eq. (20). Then we start at L₀ and draw a line L₀Δ to locate V₁. Then an equilibrium tie line through V₁ locates L₁. Line L₁Δ is drawn to give V₂. A tie line from V₂ gives L₂. This is continued until the desired L_N is reached.

Example E6: Number of stages in countercurrent extraction

Pure isopropyl ether (C) of 450 kg/h is being used to extract an aqueous solution of 150 kg/h with 30 wt % acetic acid (A) and 70 wt % water (B) by countercurrent multistage extraction. The exit acid concentration in the aqueous phase is 10 wt %. Calculate the number of stages required.

Solution: draw a diagram,

Known values:

Pure solvent from N+1: V_N+1 = 450 kg/h y_A,N+1 = y_B,N+1 = 0 yC,N+1 = 1.0, Feed: L₀ = 150 kg/h x_A0 = 0.3 x_B0 = 0.7 x_C0 = 0

Exit in water phase: x_AN = 0.1.

L_N must be in the raffinate solubility line. The points V_N+1, L₀, and L_N are plotted.

x_A, y_A 0.0 0.1 0.2 0.3 0.4 0.5 x C , y C 0.0 0.2 0.4 0.6 0.8 1.0 A (acetic acid) B (water) C (isopropyl ether) L₀ L_N V₁ V_N+1

The mixture points are found by Eqs. (13) & (14), x_CM = 0.75, x_AM = 0.075. The point V₁ is located as the intersection of line L_NM with the phase boundary in the extract phase, y_A1 = 0.072, y_C1 = 0.895.

Then lines L₀V₁, and L_NV_N+1 is drawn to locate the point Δ. x_A, y_A -0.1 0.0 0.1 0.2 0.3 0.4 0.5 x C , y C 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 A (acetic acid) B (water) C (isopropyl ether) L₀ L_N V₁ V_N+1

Δ

Starting at L₀ we draw a line L₀Δ to locate V₁. Then an equilibrium tie line through V₁ locates L₁. Line L₁Δ is drawn to give V₂. A tie line from V₂ gives L₂. A final tie line gives L₃, which is beyond the desired L_N. Hence, about 2.5 theoretical stages are needed.

3. McCabe-Thiele method

Stepping off many stages on a triangular diagram can be difficult and inaccurate. More accurate calculations can be done with a McCabe-Thiele diagram. Here we focus on the concentration of solute in the extract and raffinate phases. The diagram does not show the concentration of the diluents in the extract or the concentration of solvent in the raffinate. These minor components of both phases are accounted for in determining the total flow of extract and raffinate, which affects the position of the operating line.

In the McCabe-Thiele diagram, the equilibrium data are shown on a rectangular graph, where the mass fraction of solute in the extract (V) phase, yA, is plotted as the ordinate and the mass fraction of solute in the raffinate (L) phase, xA, as the abscissa. The conversion of equilibrium tie line in the triangle diagram to the y-x digram is shown below.

Since the total flow rates are not constant, the triangular diagram and the Δ point are used to plot a curved operating line on the McCabe-Thiele diagram. This construction is illustrated in the following figure for a single point.

The two end points of the operating line are already given in example E6. (x_AN = 0.1, y_A,N+1 = 0), and (x_A0 = 0.3, y_A1 = 0.072).

For any arbitrary operating line (must go through Δ), the values of the extract and raffinate concentrations of A are determined from the phase diagram using the common point Δ and transferred to the y-x diagram, as shown in the following figure.

The number of stages is then calculated by stepping off the triangles with the operating and equilibrium lines, which is about 2 in this case.

4. Minimum solvent rate

If a solvent rate V_N+1 is selected at too low a value, a limiting case will be reached with an operating line through Δ and a tie line being the same. Then an infinite number of stages will be needed to reach the desired separation. The minimum amount of solvent is reached. For actual operation a greater amount of solvent must be used.

The procedure to obtain this minimum solvent rate is as follows and shown in the right figure. Firstly line L_NV_N+1 is extended, then all tie lines between L₀ and L_N are drawn to intersect the extended line LNVN+1. The intersection farthest from V_N+1 (if Δ is in the L_N side, which is the case in the figure) or nearest V_N+1 (if Δ is on the V_N+1 side) is the Δ_min point for minimum solvent. The actual position of Δ must be farther from V_N+1

(if on the L_N side) or nearer to V_N+1 (if on the V_N+1 side) for a finite number of stages. The larger the amount of

Minimum solvent for countercurrent extraction.

This is proved as below. If Δ is in the L_N side,

LN - VN+1 = Δ LN = VN+1 + Δ

According to lever arm’s rule: V_N V_N L_N ΔΔL_N =( LN - VN+1) ΔLN = LN ΔLN - VN+1ΔLN V_N LNΔLN V_N L_N ΔL_N = LN VN LN LN

Since LN and VN LN are constant, VN+1 will achieve its maximum when the length of ΔL_N is at its maximum.

If Δ is in the VN+1 side,

V

V

Δ

L

L

Δ

V

L

L

Δ

V

Δ

L

L

V

V

Δ

V

Δ

=

L

1

Since LN and VN LN are constant, VN+1 will achieve its maximum when the length of V_N Δ is at its minimum.